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Suppose that \(P(A)=.4\) and \(P(B)=.2\). If events \(A\) and \(B\) are independent, find these probabilities: a. \(P(A \cap B)\) b. \(P(A \cup B)\)

Short Answer

Expert verified
Question: Given that the events A and B are independent with probabilities P(A) = 0.4 and P(B) = 0.2, find the probabilities of their intersection (A ∩ B) and union (A ∪ B). Answer: a. The probability of their intersection P(A ∩ B) is 0.08. b. The probability of their union P(A ∪ B) is 0.52.

Step by step solution

01

a. Find the probability of \(P(A \cap B)\)

As A and B are independent, their intersection probability is the product of their probabilities: \(P(A \cap B) = P(A) \cdot P(B) = 0.4 \cdot 0.2 = 0.08\)
02

b. Find the probability of \(P(A \cup B)\)

To find the probability of the union of events A and B, we will use the formula: \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) Now, we will use the values of the probabilities previously found: \(P(A \cup B) = 0.4 + 0.2 - 0.08 = 0.52\) So, the probabilities asked in the exercise are: a. \(P(A \cap B) = 0.08\) b. \(P(A \cup B) = 0.52\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Independent Events
In probability theory, the concept of independent events is crucial. When we say two events, say event A and event B, are independent, it means the occurrence of one event does not affect the probability of the other event happening. In simple terms, knowing that event A has occurred gives no information about the probability of event B. An everyday example is flipping a coin in one hand and rolling a dice in the other. The result of the coin flip doesn't change the result of the dice roll.

For two events to be considered independent, the following condition must be satisfied:
  • The multiplication rule: The probability of both events A and B occurring together (intersection) is the product of their individual probabilities: \[ P(A \cap B) = P(A) \times P(B) \]
TThis rule makes problems much simpler, as seen in the exercise above, where the independence of A and B allows us to find the intersection probability by multiplying their individual probabilities together.
Intersection Probability
When tackling probability problems, understanding the potential for events to intersect is key. The intersection of two events, often expressed as \( A \cap B \), refers to both events occurring at the same time. It signifies the joint probability where the outcomes are part of both event A and event B.

We often deal with intersections in scenarios where we want to know how likely it is for two specific conditions to be met simultaneously. For instance, finding the probability that it is both raining and a weekday would be an intersection of the event 'it's raining' and the event 'it's a weekday.'For independent events, the intersection probability can be calculated using the formula:- \[ P(A \cap B) = P(A) \times P(B) \]This formula makes it super easy, as you just need to multiply the probabilities of each event. In our exercise, this is precisely how we found out that \( P(A \cap B) = 0.08 \).
Union Probability
Union probability is perfect for situations where you want one event or another to occur. It encompasses the probability of either event A, event B, or both occurring. We symbolize this operation as \( A \cup B \).

The formula to determine the union probability is a real lifesaver when dealing with overlapping events as it helps account for the double-counted intersection:
  • \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
This formula ensures we don't double-count the common outcomes of events A and B. For example, to determine the probability of it either raining or being a weekday (or both), you subtract the probability of the intersection if there's an overlap. In the given problem, we find \( P(A \cup B) = 0.52 \), which covers all possibilities: just A, just B, and both A and B together.
Probability Theory
Probability theory is a mathematical framework that allows us to quantify uncertainty. It provides tools and principles for handling situations where we are not certain about outcomes. Probabilities range between 0 and 1, where 0 means an event will not occur, and 1 means it certainly will.

Some fundamental concepts include:
  • **Events**: These are outcomes or sets of outcomes from an experiment.
  • **Sample Space**: Represents all possible outcomes of an experiment.
  • **Probability**: A measure assigned to each event that quantifies its likelihood.
Probability theory helps solve everyday problems—from assessing risks and making decisions to modeling complex systems in science and engineering. By understanding independent events, intersection probability, and union probability, one gains a toolkit to analyze various scenarios, as demonstrated in the exercise when computing \( P(A \cap B) \) and \( P(A \cup B) \). It is a versatile and powerful tool that forms the backbone of statistical analysis.

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Most popular questions from this chapter

Many companies are now testing prospective employees for drug use. However, opponents claim that this procedure is unfair because the tests themselves are not \(100 \%\) reliable. Suppose a company uses a test that is \(98 \%\) accurate- that is, it correctly identifies a person as a drug user or nonuser with probability \(.98-\) and to reduce the chance of error, each job applicant is required to take two tests. If the outcomes of the two tests on the same person are independent events, what are the probabilities of these events? a. A nonuser fails both tests. b. A drug user is detected (i.e., he or she fails at least one test). c. A drug user passes both tests.

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You have three groups of distinctly different items, four in the first group, seven in the second, and three in the third. If you select one item from each group, how many different triplets can you form?

Under the "no pass, no play" rule for athletes, an athlete who fails a course is disqualified from participating in sports activities during the next grading period. Suppose the probability that an athlete who has not previously been disqualified will be disqualified is .15 and the probability that an athlete who has been disqualified will be disqualified again in the next time period is \(.5 .\) If \(30 \%\) of the athletes have been disqualified before, what is the unconditional probability that an athlete will be disqualified during the next grading period?

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